There may be better ways to do this, using results from Galois theory. But if you want a bare-hands method, try this. Define a map $f:\mathbb{Z}[\sqrt{-19}] \to \mathbb{Z}/2\mathbb{Z}$ by $f(a+b\sqrt{-19}) = a-b \pmod2.$ Show that $f$ is a ring homomorphism with kernel $(2, 1+\sqrt{-19}).$ Since the range of $f$ is a simple ring, it follows that the kernel must be a maximal ideal.Loiz said:hello all of you
Why is (2, 1+sqrt(-19)) a maximal ideal of Z[sqrt(-19)]? How to show such thing?
I thank you for any hints
Opalg said:There may be better ways to do this, using results from Galois theory. But if you want a bare-hands method, try this. Define a map $f:\mathbb{Z}[\sqrt{-19}] \to \mathbb{Z}/2\mathbb{Z}$ by $f(a+b\sqrt{-19}) = a-b \pmod2.$ Show that $f$ is a ring homomorphism with kernel $(2, 1+\sqrt{-19}).$ Since the range of $f$ is a simple ring, it follows that the kernel must be a maximal ideal.
The way I approached this problem was to start by writing an element $a+b\sqrt{-19}\in \mathbb{Z}[\sqrt{-19}]$ as $a+b\sqrt{-19} = (a-b) + b(1+\sqrt{-19}).$ If $a-b$ is even, say $a-b=2k$, then $a+b\sqrt{-19} = 2k + b(1+\sqrt{-19}).$ That is a linear combination of $2$ and $1+\sqrt{-19}$, and is therefore in the ideal.Loiz said:Thank you very much for your help.
but i get stuck with this
kernel($f$)= {$a + b\sqrt{-19} : f(a+b\sqrt{-19})=0$} = {$a+b\sqrt{-19} : a -b \pmod2 $}={$a+b\sqrt{-19} : a=b \pmod2 $}
I am missing something crucial here cause i just don't understand why this is equivalent to the ideal $(2, 1+\sqrt{-19})$. What does an element in this ideal look like?
Opalg said:The way I approached this problem was to start by writing an element $a+b\sqrt{-19}\in \mathbb{Z}[\sqrt{-19}]$ as $a+b\sqrt{-19} = (a-b) + b(1+\sqrt{-19}).$ If $a-b$ is even, say $a-b=2k$, then $a+b\sqrt{-19} = 2k + b(1+\sqrt{-19}).$ That is a linear combination of $2$ and $1+\sqrt{-19}$, and is therefore in the ideal.
Conversely, the set $J = \{a+b\sqrt{-19}\in \mathbb{Z}[\sqrt{-19}] : a-b \text{ is even}\}$ is an ideal in $\mathbb{Z}[\sqrt{-19}]$, which contains $2$ and $1+\sqrt{-19}$. To see that it is an ideal, you need to show that if $a+b\sqrt{-19}\in J$ and $x+y\sqrt{-19} \in \mathbb{Z}[\sqrt{-19}]$, then their product is in $J$. The calculation for that goes like this: $(a+b\sqrt{-19})(x+y\sqrt{-19}) = (ax-19by) + (ay+bx)\sqrt{-19})$, and $(ax-19by) - (ay+bx) = (a-b)x - (a+19b)y$. Since $a+19b = (a-b) + 20b$, which is an even number, it follows that $(ax-19by) - (ay+bx)$ is even, as required.
That shows that $J = (2, 1+\sqrt{-19})$. So to see whether an element $a+b\sqrt{-19}$ is in $(2, 1+\sqrt{-19})$, you just need to check whether $a-b$ is even or odd. That is what suggested to me the idea of defining the homomorphism $f:\mathbb{Z}[\sqrt{-19}] \to \mathbb{Z}/2\mathbb{Z}.$
A maximal ideal in a ring is an ideal that is not properly contained in any other ideal except for the whole ring itself. In other words, there are no larger ideals that contain the maximal ideal.
Z[sqrt(-19)] is the ring of Gaussian integers, which is formed by adding the imaginary number sqrt(-19) to the set of integers. It is also known as the Eisenstein integers.
To show that (2, 1+sqrt(-19)) is a maximal ideal, we need to prove that it is not properly contained in any other ideal. This can be done by showing that any element outside of (2, 1+sqrt(-19)) cannot generate the ideal. By using the properties of Gaussian integers, we can show that this is true.
(2, 1+sqrt(-19)) works in Z[sqrt(-19)] by acting as a filter for elements in the ring. Any element that is not a multiple of 2 or cannot be written in the form 2x + (1+sqrt(-19))y, where x and y are Gaussian integers, will not be contained in this ideal.
The ideal (2, 1+sqrt(-19)) is significant because it is the only maximal ideal in Z[sqrt(-19)]. This means that it plays a key role in the structure of the ring and can be used to understand the properties of other ideals in Z[sqrt(-19)].